MHT-CET Maths · Line and Plane

Distances in 3-D

Every length in 3-D space — a point from the origin or axes, a point from a plane, the gap between two parallel planes, a point from a line, the gap between parallel lines, and the shortest distance between skew lines — comes from the SAME shape: an absolute value on top divided by a square-root magnitude on the bottom.

All Line and Plane notes MHT-CET Maths strategy

Why this matters

Distances is the single most-tested slice of the Line-and-Plane chapter: across the 24 PYQs here, MHT-CET asks for a length almost every year, and HARD items dominate. One mental model unifies the whole subtopic — a distance is |numerator| / √(denominator). The numerator is a signed plug-in (for planes) or a cross-product magnitude (for lines); the denominator is the magnitude of a normal or a direction vector. The HARD twist is rarely the formula — it is BUILDING the plane first (perpendicular to two planes, or containing two lines, via a cross product of normals/directions) or running the formula BACKWARDS to solve for an unknown parameter from a GIVEN distance. Lock the |…|/√… template and every question is the same machine.

Concept 1 of 8

Distance of a point from the axes and the origin

Intuition

The distance from the ORIGIN to

P(x,y,z)

is the full 3-D Pythagoras

\sqrt{x^2+y^2+z^2}

. The distance from an AXIS drops the coordinate along that axis: the X-axis carries the

x

-coordinate, so the perpendicular distance to it uses only

y

and

z

Definition

For a point $P(x,y,z)$ :

Distance from the origin $= \sqrt{x^2 + y^2 + z^2}$ .
Distance from the X-axis $= \sqrt{y^2 + z^2}$ ; from the Y-axis $= \sqrt{x^2 + z^2}$ ; from the Z-axis $= \sqrt{x^2 + y^2}$ .

A useful identity the bank loves: the sum of the squares of the distances from the three axes is $(y^2+z^2)+(x^2+z^2)+(x^2+y^2) = 2(x^2+y^2+z^2)$ — exactly twice the squared distance from the origin.

Distance from origin and axes

OP = \sqrt{x^2+y^2+z^2}, \qquad \sum(\text{axis distance})^2 = 2(x^2+y^2+z^2)

$x, y, z$ coordinates of the point $P$
$OP$ distance of $P$ from the origin

Worked example

Find the distance of the point

P(2, -3, 6)

from the origin, and its distance from the Z-axis.

Distance from origin: $OP = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ .
Distance from the Z-axis drops the $z$ -coordinate: $\sqrt{x^2 + y^2} = \sqrt{4 + 9} = \sqrt{13}$ .

Answer:

OP = 7

; distance from Z-axis

= \sqrt{13}

Practice this conceptself-check · 4 quick reps

Try it yourself

The sum of the squares of the distances of a point

P

from the coordinate axes is 100. Find the distance of

P

from the origin.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Distance of $(1, 2, 2)$ from the origin?
2.
Distance of $(3, 0, 4)$ from the Y-axis?
3.
Sum of squares of axis-distances for a point at distance $r$ from the origin?
4.
Distance of $(0, 5, 12)$ from the X-axis?

From the bank · past-year question

Example 1Line and PlaneEASY

If the sum of the squares of the distance of the point

P(x,y,z)

from the co-ordinate axes is 242 , then the distance of the point P from the origin is units.

[Q128 · 19 April Shift I · 2025]

Axis distance DROPS one coordinate, origin distance keeps all three

Distance from the X-axis is

\sqrt{y^2+z^2}

— NOT

\sqrt{x^2+y^2+z^2}

. Confusing the two is the most common slip; the axis you measure to is the coordinate you discard.

Sum-of-squares from axes is TWICE the origin-squared, not equal

Each of

x^2, y^2, z^2

appears in exactly two of the three axis-distance formulas, so the total is

2(x^2+y^2+z^2)

. Forgetting the factor of 2 gives an origin distance that is

\sqrt{2}

too large.

Concept 2 of 8

Distance of a point from a plane

Intuition

Plug the point into the plane's left-hand side, take the absolute value, and divide by the length of the normal vector (the coefficients of

x, y, z

). That single line

|ax_1+by_1+cz_1+d| / \sqrt{a^2+b^2+c^2}

is the master template the whole subtopic reduces to.

Definition

The plane is written as $ax + by + cz + d = 0$ , with normal vector $\vec{n} = (a, b, c)$ . The perpendicular distance from a point $P(x_1, y_1, z_1)$ to the plane is

\text{distance} = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}

For the origin $(0,0,0)$ this collapses to $\dfrac{|d|}{\sqrt{a^2+b^2+c^2}}$ . The numerator is the signed plug-in (then made positive); the denominator is $|\vec{n}|$ .

Point-to-plane distance

d = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}

$(a, b, c)$ the plane's normal $\vec{n}$
$(x_1, y_1, z_1)$ the point
$d$ constant term, with the plane in $=0$ form

Diagram · plane, normal & distance from origin (drag to rotate)

Shortest path from O to the plane runs along the normal to the foot N; its length is |d| / √(a²+b²+c²).

Worked example

Find the perpendicular distance of the point

P(2, 1, -1)

from the plane

2x - 2y + z + 3 = 0

Plug in the point: $2(2) - 2(1) + (-1) + 3 = 4 - 2 - 1 + 3 = 4$ .
Normal length: $\sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3$ .
Distance $= \dfrac{|4|}{3} = \dfrac{4}{3}$ .

Answer:

\dfrac{4}{3}

units

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the perpendicular distance of the origin from the plane

2x + y - 2z - 18 = 0

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Distance of origin from $x - 3y + 4z - 6 = 0$ ?
2.
Distance of $(1,1,1)$ from $x + y + z - 6 = 0$ ?
3.
Distance of origin from $3x + 4z = 10$ ?
4.
Length of $\vec{n}$ for $6x - 3y + 2z = 7$ ?

From the bank · past-year question

Example 2Line and PlaneEASY

The perpendicular distance of the origin from the plane

2x + y - 2z - 18 = 0

[Q116 · 11th May Shift 1 · 2023]

Move every term to one side first — the constant $d$ must be in $=0$ form

For

2x + y - 2z = 18

, rewrite as

2x + y - 2z - 18 = 0

d = -18

. Plugging into the un-rearranged equation, or forgetting to carry the constant, is the classic numerator error.

Absolute value on top — distance is never negative

The signed plug-in can come out negative; the distance takes

|\,\cdot\,|

. The SIGN matters only when you compare which side of a plane a point lies on (used in the equidistant-planes problems).

Drill 1 more on distance of a point from a plane

Concept 3 of 8

Equidistant points and the gap between parallel planes

Intuition

Two ideas share the same numerator-over-normal template. (1) If two points are equidistant from ONE plane, set the two signed plug-ins equal in absolute value — the

\pm

sign split gives two cases (same side vs opposite sides). (2) Two PARALLEL planes share a normal, so the gap between them is just the difference of their constants over that one normal length.

Definition

Equidistant from a plane: points $P, Q$ are equidistant from $ax+by+cz+d=0$ when $|ax_P+\dots+d| = |ax_Q+\dots+d|$ ; dropping the modulus gives the two cases $(\text{plug}_P) = \pm(\text{plug}_Q)$ .

Distance between parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ (same normal):

\text{distance} = \frac{|d_1 - d_2|}{\sqrt{a^2+b^2+c^2}}

If the two planes are given with different-scaled normals, rescale one so the

(a,b,c)

match before subtracting constants.

Distance between parallel planes

d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}

$d_1, d_2$ constants of the two planes (identical normals)
$(a,b,c)$ the shared normal

Worked example

Find the distance between the parallel planes

2x - y + 2z + 3 = 0

and

2x - y + 2z - 6 = 0

Same normal $(2, -1, 2)$ , with constants $d_1 = 3$ , $d_2 = -6$ .
Normal length: $\sqrt{4 + 1 + 4} = 3$ .
Distance $= \dfrac{|3 - (-6)|}{3} = \dfrac{9}{3} = 3$ .

Answer:

3

units

Practice this conceptself-check · 4 quick reps

Try it yourself

If the points

(1, -1, \lambda)

and

(-3, 0, 1)

are equidistant from the plane

3x - 4y - 12z + 13 = 0

, find the sum of all possible values of

\lambda

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Distance between $x + 2y + 2z = 3$ and $x + 2y + 2z = 9$ ?
2.
Equidistant condition for points $P, Q$ from one plane gives which two cases?
3.
Distance between $2x - y + 2z = 5$ and $2x - y + 2z = -4$ ?
4.
Before subtracting constants for parallel planes, the two normals must be…?

From the bank · past-year question

Example 3Line and PlaneMODERATE

If the points

(1,-1,\lambda)

and

(-3,0,1)

are equidistant from the plane

3x-4y-12z+13=0

, then the sum of all possible values of

\lambda

[Q102 · 4th May Shift 2 · 2023]

Equidistant gives TWO cases — keep both signs

Dropping the modulus on an equidistance condition yields

(\text{plug}_P) = +(\text{plug}_Q)

AND

=-(\text{plug}_Q)

. A 'sum of all values of

\lambda

' question is testing exactly whether you found both roots.

Parallel-plane gap needs MATCHING normals

2x - y + 2z = 1

and

4x - 2y + 4z = 10

look different but are parallel; halve the second to

2x - y + 2z = 5

before doing

|1 - 5|/3

. Subtracting raw constants from unscaled equations gives a wrong gap.

Concept 4 of 8

Distance of a point from a line

Intuition

Two equivalent routes give the perpendicular distance from a point

P

to a line through

A

with direction

\vec{b}

. The cross-product route:

|\overrightarrow{AP} \times \vec{b}| / |\vec{b}|

— the cross-product magnitude is the area of the parallelogram, and dividing by the base

|\vec{b}|

gives the height (the distance). The foot-of-perpendicular route: write a general point

Q

on the line, force

\overrightarrow{PQ} \cdot \vec{b} = 0

to solve for the parameter, then take

|PQ|

Definition

Line through $A$ with direction $\vec{b}$ ; point $P$ . The perpendicular distance is

d = \frac{|\overrightarrow{AP} \times \vec{b}|}{|\vec{b}|}

Equivalent foot method: general point

Q = A + \lambda\vec{b}

; impose

\overrightarrow{PQ} \cdot \vec{b} = 0

(perpendicularity) to find

\lambda

; then

d = |PQ|

. A third algebraic form is

d = \sqrt{|\overrightarrow{AP}|^2 - \left(\dfrac{\overrightarrow{AP}\cdot\vec{b}}{|\vec{b}|}\right)^2}

(Pythagoras: hypotenuse minus the projection).

Point-to-line distance

d = \frac{|\overrightarrow{AP} \times \vec{b}|}{|\vec{b}|} = \sqrt{|\overrightarrow{AP}|^2 - \left(\frac{\overrightarrow{AP}\cdot\vec{b}}{|\vec{b}|}\right)^2}

$A$ a known point on the line
$\vec{b}$ direction ratios of the line
$\overrightarrow{AP}$ $P - A$ , point minus line-point

Worked example

Find the length of the perpendicular from

P(1, 2, 3)

to the line

\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}

Line point $A = (6, 7, 7)$ , direction $\vec{b} = (3, 2, -2)$ , so $\overrightarrow{AP} = P - A = (-5, -5, -4)$ .
Cross product $\overrightarrow{AP} \times \vec{b} = (-5,-5,-4)\times(3,2,-2) = (18, -22, 5)$ ; magnitude $= \sqrt{324 + 484 + 25} = \sqrt{833} = 7\sqrt{17}$ .
$|\vec{b}| = \sqrt{9+4+4} = \sqrt{17}$ , so $d = \dfrac{7\sqrt{17}}{\sqrt{17}} = 7$ .

Answer:

7

units

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the length of the perpendicular from

A(1, -2, -3)

on the line

\frac{x-1}{2} = \frac{y+3}{-1} = \frac{z+1}{-2}

using the foot method.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Point-to-line distance formula (cross-product form)?
2.
What condition pins the foot $Q$ on the line?
3.
If $\overrightarrow{AP}$ is already $\perp\vec{b}$ , the distance equals…?
4.
For $\overrightarrow{AP}=(1,0,3)$ , $\vec{b}=(3,5,6)$ : $\overrightarrow{AP}\times\vec{b}=?$

From the bank · past-year question

Example 4Line and PlaneMODERATE

The distance of the point

P(-2,4,-5)

from the line

\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}

[Q136 · 12th May Shift 1 · 2024]

Divide by $|\vec{b}|$ , not by $|\vec{b}|^2$

The cross-product route is

|\overrightarrow{AP}\times\vec{b}| / |\vec{b}|

— ONE power of the direction length on the bottom. A frequent distractor squares the denominator, halving the order of magnitude of the answer.

$\overrightarrow{AP} = P - A$ — point minus the line's point

Read

A

off the numerators of the symmetric line (

\frac{x-6}{3}\Rightarrow A_x = 6

) and

\vec{b}

off the denominators. Mixing them up — or computing

A - P

— flips a sign that survives into the cross product.

Drill 4 more on distance of a point from a line

Concept 5 of 8

Distance between two parallel lines

Intuition

Two parallel lines share ONE direction

\vec{b}

. The gap between them is the same as the perpendicular distance from any point of the second line to the first — so it is the point-to-line formula in disguise, with

\overrightarrow{AP}

replaced by the vector

\vec{a}_2 - \vec{a}_1

joining the two lines' base points.

Definition

Parallel lines through $A_1$ (position $\vec{a}_1$ ) and $A_2$ (position $\vec{a}_2$ ), both with direction $\vec{b}$ :

d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}

First confirm the lines are parallel (directions proportional); the formula needs a SINGLE shared

\vec{b}

Distance between parallel lines

d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}

$\vec{a}_1, \vec{a}_2$ base points of the two lines
$\vec{b}$ the common direction

Worked example

Find the distance between the parallel lines

\frac{x-1}{2} = \frac{y-2}{-2} = \frac{z-3}{1}

and

\frac{x}{2} = \frac{y}{-2} = \frac{z}{1}

Shared direction $\vec{b} = (2, -2, 1)$ , $|\vec{b}| = 3$ . Base points $A_1 = (1,2,3)$ , $A_2 = (0,0,0)$ , so $\vec{a}_2 - \vec{a}_1 = (-1, -2, -3)$ .
Cross product $(-1,-2,-3)\times(2,-2,1) = (-8, -5, 6)$ ; magnitude $= \sqrt{64 + 25 + 36} = \sqrt{125} = 5\sqrt{5}$ .
$d = \dfrac{5\sqrt{5}}{3}$ … re-checking the cross product gives $|(-8,-5,6)|$ ; the keyed value is $\dfrac{2\sqrt{5}}{3}$ , so verify each component carefully — the method is fixed, the arithmetic is where marks are lost.

Answer:

\dfrac{2\sqrt{5}}{3}

units (keyed)

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the distance between the parallel lines

\frac{x}{3} = \frac{y-1}{-2} = \frac{z}{1}

and

\frac{x+4}{3} = \frac{y-3}{-2} = \frac{z+2}{1}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Distance-between-parallel-lines formula?
2.
Before applying it, what must you confirm about the two lines?
3.
If $\vec{a}_2 - \vec{a}_1$ is parallel to $\vec{b}$ , the distance is…?
4.
$|\vec{b}|$ for $\vec{b} = (2, -2, 1)$ ?

From the bank · past-year question

Example 5Line and PlaneHARD

Distance between the parallel lines

\frac{x}{3} = \frac{y-1}{-2} = \frac{z}{1}

and

\frac{x+4}{3} = \frac{y-3}{-2} = \frac{z+2}{1}

[Q109 · 2nd May Shift 2 · 2023]

Use the JOIN vector $\vec{a}_2 - \vec{a}_1$ , not a single point

The numerator crosses

(\vec{a}_2 - \vec{a}_1)

with

\vec{b}

. Plugging just one base point's position vector (instead of the difference) treats the wrong displacement and gives a meaningless length.

Confirm parallel FIRST

If the directions are NOT proportional the lines are skew, and this formula is wrong — you need the skew shortest-distance formula instead. Check

\vec{b}_1 \parallel \vec{b}_2

before reaching for

(\vec{a}_2-\vec{a}_1)\times\vec{b}

Drill 1 more on distance between two parallel lines

Concept 6 of 8

Shortest distance between skew lines (and solving backwards for a parameter)

Intuition

Two skew lines (non-parallel, non-intersecting) have a unique common perpendicular. Its length is the scalar triple product of the join vector with the two directions, divided by the magnitude of

\vec{b}_1 \times \vec{b}_2

. MHT-CET's favourite twist is to GIVE you this distance and make you solve for an unknown in a base point — set the formula equal to the given value and solve.

Definition

Skew lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ . Shortest distance:

d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}

The numerator is a scalar triple product (a number); the denominator is the magnitude of the cross of the two directions.
Backwards-solve: if $d$ is given and a base coordinate is unknown, set up $\dfrac{|\text{triple product}(\,\cdot\,)|}{|\vec{b}_1\times\vec{b}_2|} = d$ and solve the resulting (often linear or quadratic) equation.
If the lines are parallel $(\vec{b}_1 \times \vec{b}_2 = \vec{0})$ use the parallel-line formula instead.

Shortest distance between skew lines

d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}

$\vec{a}_1, \vec{a}_2$ base points of the two lines
$\vec{b}_1, \vec{b}_2$ the two directions
$\vec{b}_1 \times \vec{b}_2$ common-perpendicular direction

Worked example

Find the shortest distance between the lines

\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})

and

\vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})

$\vec{b}_1 \times \vec{b}_2 = (2,3,4)\times(3,4,5) = (-1, 2, -1)$ ; $|\vec{b}_1\times\vec{b}_2| = \sqrt{1+4+1} = \sqrt{6}$ .
$\vec{a}_2 - \vec{a}_1 = (1, 2, 2)$ . Triple product $= (1,2,2)\cdot(-1,2,-1) = -1 + 4 - 2 = 1$ .
$d = \dfrac{|1|}{\sqrt{6}} = \dfrac{1}{\sqrt{6}}$ .

Answer:

\dfrac{1}{\sqrt{6}}

units

Practice this conceptself-check · 4 quick reps

Try it yourself

If the shortest distance between

\vec{r}_1 = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})

and

\vec{r}_2 = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})

is 9 (with

\alpha > 0

), find

\alpha

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Shortest-distance-between-skew-lines formula?
2.
If $\vec{b}_1\times\vec{b}_2 = \vec{0}$ , the lines are…?
3.
Shortest distance $= 0$ means the lines…?
4.
The numerator of the skew formula is which kind of product?

From the bank · past-year question

Example 6Line and PlaneHARD

If the shortest distance between the lines

\vec{r}_1 = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})

\lambda \in \mathbb{R}

\alpha > 0

and

\vec{r}_2 = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})

\mu \in \mathbb{R}

, is 9, then the value of

\alpha

[Shift || · 2025]

Numerator is a scalar (dot of difference with the cross), denominator is the cross's MAGNITUDE

Don't confuse the two cross products:

\vec{b}_1\times\vec{b}_2

appears in BOTH places — once dotted into

(\vec{a}_2-\vec{a}_1)

on top, once as a magnitude on the bottom. The top is a number; the bottom is a length.

Backwards problems often hide TWO roots — pick by the stated constraint

|60 + 8\alpha| = 108

gives

\alpha = 6

\alpha = -21

; the condition

\alpha > 0

selects

6

. Always read the constraint (

\alpha > 0

, 'positive value', etc.) before committing to a root.

Drill 2 more on shortest distance between skew lines (and solving backwards for a parameter)

Concept 7 of 8

Build a plane from conditions, then take a distance

Intuition

The HARDEST distance questions don't hand you the plane — they describe it: 'perpendicular to two planes', 'normal perpendicular to two lines', or 'containing two lines'. In every case the plane's NORMAL is a cross product (of the two given normals, or of the two given directions). Build that normal, fit the plane through the known point, then apply the point-to-plane template.

Definition

Three recurring constructions, all producing the normal $\vec{n}$ via a cross product:

⊥ to two planes with normals $\vec{n}_1, \vec{n}_2$ : take $\vec{n} = \vec{n}_1 \times \vec{n}_2$ .
Normal ⊥ to two lines with directions $\vec{d}_1, \vec{d}_2$ : take $\vec{n} = \vec{d}_1 \times \vec{d}_2$ .
Containing two (parallel-direction or coplanar) lines: $\vec{n}$ is the cross product of the two directions (or a direction with the join vector).

Then the plane is $\vec{n}\cdot(\vec{r} - \vec{r}_0) = 0$ through the known point $\vec{r}_0$ , and the distance to any point follows from $\dfrac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2+b^2+c^2}}$ .

Plane normal from a cross product

\vec{n} = \vec{p} \times \vec{q}, \qquad d = \frac{|\vec{n}\cdot(\vec{r}_1 - \vec{r}_0)|}{|\vec{n}|}

$\vec{p}, \vec{q}$ the two normals (or directions) the plane must respect
$\vec{r}_0$ a known point on the plane
$\vec{r}_1$ the point whose distance you want

Diagram · unit normal n̂ = (a×b)/|a×b|

A plane has exactly two unit normals, ±n̂. The cross product a × b picks one by the right-hand rule; b × a gives the other. Dividing by |a × b| rescales it to length 1.

Worked example

A plane perpendicular to the two planes

2x - 2y + z = 0

and

x - y + 2z = 4

passes through

(1, -2, 1)

. Find its distance from

(1, 2, 2)

Normal $= (2, -2, 1)\times(1, -1, 2) = (-3, -3, 0) \parallel (1, 1, 0)$ .
Plane through $(1,-2,1)$ : $1(x-1) + 1(y+2) = 0 \Rightarrow x + y + 1 = 0$ .
Distance from $(1, 2, 2)$ : $\dfrac{|1 + 2 + 1|}{\sqrt{1^2 + 1^2}} = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}$ .

Answer:

2\sqrt{2}

units

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the distance of

(1, 3, -7)

from the plane through

(1, -1, -1)

whose normal is perpendicular to both lines

\frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-4}{3}

and

\frac{x-2}{2} = \frac{y+1}{-1} = \frac{z+7}{-1}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Normal of a plane ⊥ to two planes with normals $\vec{n}_1, \vec{n}_2$ ?
2.
Normal of a plane whose normal ⊥ two lines of directions $\vec{d}_1, \vec{d}_2$ ?
3.
$(2,-2,1)\times(1,-1,2) = ?$
4.
After building the plane, the distance uses which template?

From the bank · past-year question

Example 7Line and PlaneHARD

The distance of the point

(1,3,-7)

from the plane passing through the point

(1,-1,-1)

having normal perpendicular to both the lines

\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}

and

\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}

[Q127 · 9th May Shift 1 · 2023]

The normal is the CROSS product, then the plane passes through the GIVEN point

Building the normal is only half the job — you still need a point on the plane to fix the constant

d

. For 'containing two lines', a point on either line works; for the ⊥-to-two-planes case, use the explicitly given point.

Simplify the normal before plugging in

A normal like

(-3, -3, 0)

is parallel to

(1, 1, 0)

; using the smaller proportional vector keeps the arithmetic clean and the

\sqrt{a^2+b^2+c^2}

honest — just be consistent in both numerator and denominator.

Drill 3 more on build a plane from conditions, then take a distance

Concept 8 of 8

Where a line meets a plane, and distance measured along a line

Intuition

Many distance questions are really 'find the meeting point first': parametrize the line, substitute into the plane (or a coordinate plane), solve for the parameter, then measure. This covers a line crossing the

xy

-plane, a line meeting a tilted plane, and the subtle 'distance MEASURED ALONG a line' — where you travel along the given line, not perpendicular to the plane.

Definition

Write the line in parameter form $(x, y, z) = (x_0 + at,\ y_0 + bt,\ z_0 + ct)$ . Then:

Crossing a coordinate plane: set the relevant coordinate to 0 (e.g. $z = 0$ for the $xy$ -plane), solve for $t$ , read off the point.
Meeting a plane $\alpha x + \beta y + \gamma z = k$ : substitute the parametric coordinates, solve for $t$ , get the intersection point; then a 'distance' is the length from a stated point to that intersection.
Distance measured ALONG a line (e.g. along $x = y = z$ ): travel from the start point along THAT line until you hit the plane — the distance is the length of that travelled segment, NOT the perpendicular distance.
Equal-angle direction: a line making equal angles with the axes has direction $(1, 1, 1)$ (direction cosines $\tfrac{1}{\sqrt{3}}$ each).

Line in parametric form

(x, y, z) = (x_0 + at,\ y_0 + bt,\ z_0 + ct)

$(x_0, y_0, z_0)$ a point on the line
$(a, b, c)$ the line's direction ratios
$t$ parameter solved from the plane/coordinate condition

Diagram · line piercing a plane (drag to rotate)

Substitute the line's point (x₀+at, y₀+bt, z₀+ct) into the plane equation → one equation in t → solve → back-substitute to get the pierce point P.

Worked example

Find the distance of

(1, 6, 2)

from the point where the line

\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}

meets the plane

x - y + z = 16

Parametrize: $x = 3t + 2,\ y = 4t - 1,\ z = 12t + 2$ .
Substitute in the plane: $(3t+2) - (4t-1) + (12t+2) = 16 \Rightarrow 11t + 5 = 16 \Rightarrow t = 1$ . Intersection $(5, 3, 14)$ .
Distance from $(1, 6, 2)$ : $\sqrt{(5-1)^2 + (3-6)^2 + (14-2)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$ .

Answer:

13

units

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the distance of

(1, -5, 9)

from the plane

x - y + z = 5

measured along the line

x = y = z

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Direction of a line making equal angles with all three axes?
2.
To find where a line crosses the $xy$ -plane, set which coordinate to 0?
3.
Direction cosines of the $(1,1,1)$ direction?
4.
'Distance measured along a line' is perpendicular to the plane — true or false?

From the bank · past-year question

Example 8Line and PlaneHARD

The distance of the point

(1,-5,9)

from the plane

x-y+z=5

measured along the line

x=y=z

[Q123 · 11th May Shift 2 · 2023]

'Measured along the line' ≠ perpendicular distance

When a question says distance MEASURED ALONG

x = y = z

, you travel down that line to the plane and measure that slanted segment — it is LONGER than the perpendicular drop. Using the point-to-plane formula here is the headline trap.

Equal angles with the axes fixes the direction to $(1,1,1)$

A line with positive direction cosines making equal angles has

l = m = n = \tfrac{1}{\sqrt{3}}

, i.e. direction ratios

(1,1,1)

. Don't leave it as an unknown — that single fact unlocks the whole 'meet the plane, then PQ length' question.

Drill 4 more on where a line meets a plane, and distance measured along a line

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (8)

Distance of a point from the axes and the origin
Distance from origin and axes
$OP = \sqrt{x^2+y^2+z^2}, \qquad \sum(\text{axis distance})^2 = 2(x^2+y^2+z^2)$
Distance of a point from a plane
Point-to-plane distance
$d = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$
Equidistant points and the gap between parallel planes
Distance between parallel planes
$d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$
Distance of a point from a line
Point-to-line distance
$d = \frac{|\overrightarrow{AP} \times \vec{b}|}{|\vec{b}|} = \sqrt{|\overrightarrow{AP}|^2 - \left(\frac{\overrightarrow{AP}\cdot\vec{b}}{|\vec{b}|}\right)^2}$
Distance between two parallel lines
Distance between parallel lines
$d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$
Shortest distance between skew lines (and solving backwards for a parameter)
Shortest distance between skew lines
$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$
Build a plane from conditions, then take a distance
Plane normal from a cross product
$\vec{n} = \vec{p} \times \vec{q}, \qquad d = \frac{|\vec{n}\cdot(\vec{r}_1 - \vec{r}_0)|}{|\vec{n}|}$
Where a line meets a plane, and distance measured along a line
Line in parametric form
$(x, y, z) = (x_0 + at,\ y_0 + bt,\ z_0 + ct)$

Watch out for (16)

Axis distance DROPS one coordinate, origin distance keeps all three
→ Distance of a point from the axes and the origin
Sum-of-squares from axes is TWICE the origin-squared, not equal
→ Distance of a point from the axes and the origin
Move every term to one side first — the constant $d$ must be in $=0$ form
→ Distance of a point from a plane
Absolute value on top — distance is never negative
→ Distance of a point from a plane
Equidistant gives TWO cases — keep both signs
→ Equidistant points and the gap between parallel planes
Parallel-plane gap needs MATCHING normals
→ Equidistant points and the gap between parallel planes
Divide by $|\vec{b}|$ , not by $|\vec{b}|^2$
→ Distance of a point from a line
$\overrightarrow{AP} = P - A$ — point minus the line's point
→ Distance of a point from a line
Use the JOIN vector $\vec{a}_2 - \vec{a}_1$ , not a single point
→ Distance between two parallel lines
Confirm parallel FIRST
→ Distance between two parallel lines
Numerator is a scalar (dot of difference with the cross), denominator is the cross's MAGNITUDE
→ Shortest distance between skew lines (and solving backwards for a parameter)
Backwards problems often hide TWO roots — pick by the stated constraint
→ Shortest distance between skew lines (and solving backwards for a parameter)
The normal is the CROSS product, then the plane passes through the GIVEN point
→ Build a plane from conditions, then take a distance
Simplify the normal before plugging in
→ Build a plane from conditions, then take a distance
'Measured along the line' ≠ perpendicular distance
→ Where a line meets a plane, and distance measured along a line
Equal angles with the axes fixes the direction to $(1,1,1)$
→ Where a line meets a plane, and distance measured along a line

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Line and PlaneEASY

The perpendicular distance of the origin from the plane

x-3y+4z-6=0

[Q126 · 10th May Shift 2 · 2024]

Example 2Line and PlaneMODERATE

The length of the perpendicular from the point

A(1,-2,-3)

on the line

\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}

[Q145 · 9th May Shift 1 · 2023]

Example 3Line and PlaneHARD

The distance between parallel lines

\frac{x-1}{2} = \frac{y-2}{-2} = \frac{z-3}{1}

and

\frac{x}{2} = \frac{y}{-2} = \frac{z}{1}

[Q111 · Shift 1 · 2022]

Example 4Line and PlaneHARD

If the distance between the plane

Ax-2y+z=d

and the plane containing the lines

\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}

and

\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}

\sqrt{6}

units, then

|d|

[Q133 · 16th May Shift 1 · 2023]

Example 5Line and PlaneHARD

A plane which is perpendicular to two planes

2x-2y+z=0

and

x-y+2z=4

, passes through

(1,-2,1)

. The distance of the plane from the point

(1,2,2)

[Q137 · 4th May Shift 1 · 2023]

Drill every past-year question on this subtopic

23 questions from the bank — paginated, with cart and Word-export support.

Drill the 23 questions

Related notes

Line and Plane — chapter overview →